285

## Chapter 7

### Collecting and

### Preparing Samples

### Chapter Overview

### 7A The Importance of Sampling

### 7B Designing a Sampling Plan

### 7C Implementing the Sampling Plan

### 7D Separating The Analyte From Interferents

### 7E General Theory of Separation Efficiency

### 7F Classifying Separation Techniques

### 7G Liquid –Liquid Extractions

### 7H Separation Versus Preconcentration

### 7I Key Terms

### 7J Chapter Summary

### 7K Problems

### 7L Solutions to Practice Exercises

### W

hen we use an analytical method to solve a problem, there is no guarantee that our results will be accurate or precise. In designing an analytical method we consider potential sources of determinate error and indeterminate error, and take appropriate steps to minimize their effect, such as including reagent blanks and calibrating instruments. Why might a carefully designed analytical method give poor results? One possibility is that we may have failed to account for errors associated with the sample. If we collect the wrong sample, or if we lose analyte while preparing the sample for analysis, then we introduce a determinate source of error. If we fail to collect enough samples, or if we collect samples of the wrong size, then our precision may suffer. In this chapter we consider how collecting samples and preparing them for analysis affects the accuracy and precision of our results.**7A The Importance of Sampling**

When a manufacturer lists a chemical as ACS Reagent Grade, they must demonstrate that it conforms to specifications set by the American Chemi-cal Society (ACS). For example, the ACS specifications for NaBr require that the concentration of iron be ≤5 ppm. To verify that a production lot meets this standard, the manufacturer collects and analyzes several samples, reporting the average result on the product’s label (Figure 7.1).

If the individual samples do not accurately represent the population from which they are drawn—what we call the target population—then even a careful analysis must yield an inaccurate result. Extrapolating this result from a sample to its target population introduces a determinate sam-pling error. To minimize this determinate samsam-pling error, we must collect the right sample.

Even if we collect the right sample, indeterminate sampling errors may
limit the usefulness of our analysis. Equation 7.1 shows that a confidence
interval about the mean,*X* , is proportional to the standard deviation, s, of
the analysis

µ= ±*X* *ts*

*n* 7.1

where n is the number of samples and t is a statistical factor that accounts for the probability that the confidence interval contains the true value, m.

Each step of an analysis contributes random error that affects the
over-all standard deviation. For convenience, let’s divide an analysis into two
steps—collecting the samples and analyzing the samples—each
character-ized by a standard deviation. Using a propagation of uncertainty, the
rela-tionship between the overall variance, s2_{, and the variances due to sampling, }

*s*_{samp}2 _{, and the analytical method, }* _{s}*
meth
2

_{, is }

*s*2

_{=}

*s*2

_{+}

*s*2

samp meth 7.2

Equation 7.2 shows that the overall variance for an analysis may be
limited by either the analytical method or the collecting of samples.
Unfor-tunately, analysts often try to minimize the overall variance by improving
only the method’s precision. This is a futile effort, however, if the standard
deviation for sampling is more than three times greater than that for the
method.1 _{Figure 7.2}_{ shows how the ratio }_{s}

samp/smeth affects the method’s
contribution to the overall variance. As shown by the dashed line, if the
sample’s standard deviation is 3× the method’s standard deviation, then
indeterminate method errors explain only 10% of the overall variance. If
in-determinate sampling errors are significant, decreasing s_{meth} provides only
a nominal change in the overall precision.

1 Youden, Y. J. *J. Assoc. Off. Anal. Chem.***1981**, *50*, 1007–1013.

**Figure 7.1** Certificate of
analy-sis for a production lot of NaBr.
The result for iron meets the ACS
specifications, but the result for
potassium does not.

Equation 7.1 should be familiar to you.
See Chapter 4 to review confidence
inter-vals and see Appendix 4 for values of *t*.

For a review of the propagation of
uncer-tainty, see Chapter 4C and Appendix 2.
Although equation 7.1 is written in terms
of a standard deviation, *s*, a propagation
of uncertainty is written in terms of
vari-ances, *s*2_{. In this section, and those that }

follow, we will use both standard devia-tions and variances to discuss sampling uncertainty.

### Example 7.1

A quantitative analysis gives a mean concentration of 12.6 ppm for an
analyte. The method’s standard deviation is 1.1 ppm and the standard
deviation for sampling is 2.1 ppm. (a) What is the overall variance for the
analysis? (b) By how much does the overall variance change if we improve
*s*meth by 10% to 0.99 ppm? (c) By how much does the overall variance
change if we improve s_{samp} by 10% to 1.89 ppm?

**S**

**S**

**olution**(a) The overall variance is

*s*2_{=}*s*2 _{+}*s*2 _{=} _{2 1} 2_{+} _{1 1} 2_{=}_{5 6}
samp meth ( . ppm) ( . ppm) . pppm

2

(b) Improving the method’s standard deviation changes the overall vari-ance to

*s*2_{=}_{( .}_{2 1}_{ppm}_{)}2_{+}_{( .}_{0 99}_{ppm}_{)}2_{=}_{5 4}_{.} _{ppm}2

Improving the method’s standard deviation by 10% improves the overall variance by approximately 4%.

(c) Changing the standard deviation for sampling

*s*2 _{=}_{( .}_{1 9}_{ppm}_{)}2_{+}_{( .}_{1 1}_{ppm}_{)}2_{=}_{4 8}_{.} _{ppm}2

improves the overall variance by almost 15%. As expected, because
*s*_{samp} is larger than s_{meth}, we obtain a bigger improvement in the
over-all variance when we focus our attention on sampling problems.
To determine which step has the greatest effect on the overall variance,
we need to measure both ssamp and smeth. The analysis of replicate samples

**Figure 7.2** The blue curve shows the method’s
contribu-tion to the overall variance, *s*2_{, as a function of the relative }

magnitude of the standard deviation in sampling, *s*_{samp},
and the method’s standard deviation, *s*_{meth}. The dashed
red line shows that the method accounts for only 10% of
the overall variance when *s*_{samp}= 3 × *s*_{meth}.
Understand-ing the relative importance of potential sources of
inde-terminate error is important when considering how to
improve the overall precision of the analysis.

### Practice Exercise 7.1

Suppose you wish to reduce the
overall variance in Example 7.1
to 5.0 ppm2. If you focus on the
method, by what percentage do
you need to reduce s_{meth}? If you
focus on the sampling, by what
percentage do you need to
re-duce s_{samp}?

Click here to review your answer
to this exercise
0 1 2 3 4 5
0
20
40
60
80
100
*s*samp/*s*meth
Per
cen
t of O
ver
all
Var
ianc
e D
ue t
o the M
ethod

provides an estimate of the overall variance. To determine the method’s variance we analyze samples under conditions where we may assume that the sampling variance is negligible. The sampling variance is determined by difference.

### Example 7.2

The following data were collected as part of a study to determine the effect of sampling variance on the analysis of drug-animal feed formulations.2

% Drug (w/w) % Drug (w/w)

0.0114 0.0099 0.0105 0.0105 0.0109 0.0107

0.0102 0.0106 0.0087 0.0103 0.0103 0.0104

0.0100 0.0095 0.0098 0.0101 0.0101 0.0103

0.0105 0.0095 0.0097

The data on the left were obtained under conditions where both s_{samp} and
*s*_{meth} contribute to the overall variance. The data on the right were obtained
under conditions where ssamp is known to be insignificant. Determine the
overall variance, and the standard deviations due to sampling and the
ana-lytical method. To which factor—sampling or the method—should you
turn your attention if you want to improve the precision of the analysis?

**S**

**S**

**olution**Using the data on the left, the overall variance, s2_{, is 4.71 }_{×}_{ 10}–7_{. To find }
the method’s contribution to the overall variance, *s*meth2 , we use the data on

the right, obtaining a value of 7.00 × 10–8. The variance due to sampling,

*s*_{samp}2 _{, is}

*s*_{samp}2 _{= −}*s*2 *s*_{meth}2 _{=}_{4 71 10}_{.} _{×} −7_{−}_{7 00 10}_{.} _{×} −8_{=}_{4 01 1}_{.} _{×} _{00}−7

Converting variances to standard deviations gives s_{samp} as 6.32 × 10–4 and
*s*meth as 2.65 × 10–4. Because ssamp is more than twice as large as smeth,
im-proving the precision of the sampling process has the greatest impact on
the overall precision.

2 Fricke, G. H.; Mischler, P. G.; Staffieri, F. P.; Houmyer, C. L. *Anal. Chem.***1987**, *59*, 1213–
1217.

There are several ways to minimize the standard deviation for sampling. Here are two examples. One approach is to use a standard reference material (SRM) that has been carefully prepared to minimize indeterminate sampling errors. When the sample is homogeneous—as is the case, for example, with aqueous samples—a useful approach is to conduct replicate analyses on a single sample.

See Chapter 4 for a review of how to cal-culate the variance.

### Practice Exercise 7.2

A polymer’s density provides a measure of its crystallinity. The standard deviation for the determination of density using a single sample of a poly-mer is 1.96 × 10–3 g/cm3. The standard deviation when using different samples of the polymer is 3.65 × 10–2 g/cm3. Determine the standard deviations due to sampling and the analytical method.

**7B Designing A Sampling Plan**

A sampling plan must support the goals of an analysis. For example, a material scientist interested in characterizing a metal’s surface chemistry is more likely to choose a freshly exposed surface, created by cleaving the sam-ple under vacuum, than a surface previously exposed to the atmosphere. In a qualitative analysis, a sample does not need to be identical to the original substance, provided that there is sufficient analyte to ensure its detection. In fact, if the goal of an analysis is to identify a trace-level component, it may be desirable to discriminate against major components when collect-ing samples.

For a quantitative analysis, the sample’s composition must accurately represent the target population, a requirement that necessitates a careful sampling plan. Among the issues to consider are these five questions.

1. From where within the target population should we collect samples? 2. What type of samples should we collect?

3. What is the minimum amount of sample for each analysis? 4. How many samples should we analyze?

5. How can we minimize the overall variance for the analysis?

**7B.1 Where to Sample the Target Population**

A sampling error occurs whenever a sample’s composition is not identical
to its target population. If the target population is homogeneous, then
we can collect individual samples without giving consideration to where to
sample. Unfortunately, in most situations the target population is
hetero-geneous. Due to settling, a medication available as an oral suspension may
have a higher concentration of its active ingredients at the bottom of the
container. The composition of a clinical sample, such as blood or urine, may
depend on when it is collected. A patient’s blood glucose level, for instance,
changes in response to eating and exercise. Other target populations show
both a spatial and a temporal heterogeneity. The concentration of dissolved
O_{2} in a lake is heterogeneous due both to the changing seasons and to point
sources of pollution.

If the analyte’s distribution within the target population is a concern, then our sampling plan must take this into account. When feasible, homog-enizing the target population is a simple solution—in most cases, however, this is impracticable. Additionally, homogenization destroys information about the analyte’s spatial or temporal distribution within the target popu-lation, information that may be of importance.

**Random Sampling**

The ideal sampling plan provides an unbiased estimate of the target popu-lation’s properties. A random sampling is the easiest way to satisfy this

The composition of a homogeneous
tar-get population is the same regardless of
where we sample, when we sample, or the
size of our sample. For a heterogeneous
target population, the composition is not
the same at different locations, at different
times, or for different sample sizes.
For an interesting discussion of the
impor-tance of a sampling plan, see Buger, J. et
al. “Do Scientists and Fishermen Collect
the Same Size Fish? Possible Implications
for Exposure Assessment,” *Environ. Res.*

requirement.3_{ Despite its apparent simplicity, a truly random sample} _{is }
difficult to collect. Haphazard sampling, in which samples are collected
without a sampling plan, is not random and may reflect an analyst’s
unin-tentional biases.

Here is a simple method for ensuring that we collect random samples. First, we divide the target population into equal units and assign a unique number to each unit. Then, we use a random number table to select the units to sample. Example 7.3 provides an illustrative example.

### Example 7.3

To analyze a polymer’s tensile strength, individual samples of the polymer are held between two clamps and stretched. In evaluating a production lot, the manufacturer’s sampling plan calls for collecting ten 1 cm × 1 cm samples from a 100 cm × 100 cm polymer sheet. Explain how we can use a random number table to ensure that our samples are random.

**S**

**S**

**olution**As shown by the grid, we divide the polymer sheet into 10 000 1 cm × 1 cm squares, each identified by its row number and its column number, with numbers running from 0 to 99.

For example, the blue square is in row 98 and column 1. To select ten squares at random, we enter the random num-ber table in Appendix 14 at an arbitrary point, and let the entry’s last four digits represent the row and the column for the first sample. We then move through the table in a predetermined fashion, select-ing random numbers until we have 10 samples. For our first sample, let’s use the

second entry in the third column, which is 76831. The first sample, there-fore, is row 68 and column 31. If we proceed by moving down the third column, then the 10 samples are as follows:

Sample Number Row Column Sample Number Row Column

1 76831 68 31 6 41701 17 01

2 66558 65 58 7 38605 86 05

3 33266 32 66 8 64516 45 16

4 12032 20 32 9 13015 30 15

5 14063 40 63 10 12138 21 38

In collecting a random sample we make no assumptions about the tar-get population, making this the least biased approach to sampling. On the

3 Cohen, R. D. *J. Chem. Educ.***1991**, *68*, 902–903.

Appendix 14 provides a random number table that you can use for designing sam-pling plans. 0 1 2 98 99 0 1 2 98 99

other hand, a random sample often requires more time and expense than other sampling strategies because we need a greater number of samples to ensure that we adequately sample the target population.4

**Judgmental Sampling**

The opposite of random sampling is selective, or judgmental sampling in which we use prior information about the target population to help guide our selection of samples. Judgmental sampling is more biased than random sampling, bur requires fewer samples. Judgmental sampling is useful if we wish to limit the number of independent variables influencing our results. For example, if we are studying the bioaccumulation of PCB’s in fish, we may choose to exclude fish that are too small or that appear diseased.

**SyStematic Sampling**

Random sampling and judgmental sampling represent extremes in bias and in the number of samples needed to characterize the target population. Sys-tematic sampling falls in between these extremes. In systematic sampling we sample the target population at regular intervals in space or time. Figure 7.3 shows an aerial photo of the Great Salt Lake in Utah. A railroad line divides the lake into two sections with different chemical compositions. To compare the lake’s two sections—and to evaluation spatial variations within each section—we use a two-dimensional grid to define sampling locations.

4 Borgman, L. E.; Quimby, W. F. in Keith, L. H., ed. *Principles of Environmental Sampling*,
Ameri-can Chemical Society: Washington, D. C., 1988, 25–43.

**Figure 7.3** Aerial photo of the Great Salt Lake in Utah,
taken from the International Space Station at a distance
of approximately 380 km. The railroad line divides the
lake into two sections that differ in chemical
composi-tion. Superimposing a two-dimensional grid divides each
section of the lake into sampling units. The red dots at
the center of each unit represent sampling sites. Photo
courtesy of the Image Science and Analysis Laboratory,

NASA Johnson Space Center, Photo Number

ISS007-E-13002 (eol.jsc.nasa.gov).

When a population’s heterogeneity is time-dependent, as is common in clinical studies, samples are drawn at regular intervals in time.

If a target population’s properties have a periodic trend, a systematic sampling will lead to a significant bias if our sampling frequency is too small. This is a common problem when sampling electronic signals where the problem is known as aliasing. Consider, for example, a signal consist-ing of a simple sign wave. Figure 7.4a shows how an insufficient samplconsist-ing frequency underestimates the signal’s true frequency. The apparent signal, shown by the dashed red line passing through the five data points, is signifi-cantly different from the true signal shown by the solid blue line.

According to the Nyquist theorem, to accurately determine a periodic
signal’s true frequency, we must sample the signal at least twice during each
cycle or period. If we collect samples at an interval of D*t, the highest *
fre-quency we can accurately monitor is (2Dt)–1. For example, if our sampling
rate is 1 sample/hr, the highest frequency we can monitor is (2×1 hr)–1 or
0.5 hr–1, corresponding to a period of less than 2 hr. If our signal’s period
is less than 2 hours (a frequency of more than 0.5 hr–1), then we must use
a faster sampling rate. Ideally, the sampling rate should be at least 3-4 times
greater than the highest frequency signal of interest. If our signal has a
pe-riod of one hour, we should collect a new sample every 15-20 minutes.

**SyStematic–Judgmental Sampling**

Combinations of the three primary approaches to sampling are also
pos-sible.5_{ One such combination is }_{systematic–judgmental sampling}_{, in }
which we use prior knowledge about a system to guide a systematic
sam-pling plan. For example, when monitoring waste leaching from a landfill,
we expect the plume to move in the same direction as the flow of
ground-water—this helps focus our sampling, saving money and time. The
system-atic–judgmental sampling plan in Figure 7.5 includes a rectangular grid for
most of the samples and linear transects to explore the plume’s limits.6

5 Keith, L. H. *Environ. Sci. Technol.***1990**, *24*, 610–617.

6 Flatman, G. T.; Englund, E. J.; Yfantis, A. A. in Keith, L. H., ed. *Principles of Environmental *
*Sampling*, American Chemical Society: Washington, D. C., 1988, 73–84.

**Figure 7.4** Effect of sampling frequency
when monitoring a periodic signal.
In-dividual samples are shown by the red
dots (•). In (a) the sampling frequency
is approximately 1.5 samples per period.
The dashed red line shows the apparent
signal based on five samples and the solid
blue line shows the true signal. In (b) a
sampling frequency of approximately 5
samples per period accurately reproduces
the true signal.

**Figure 7.5** Systematic–judgmental sampling plan for monitoring the leaching
of pollutants from a landfill. The sampling sites, shown as red dots (

_{•}

), are on
a systematic grid straddling the direction of the groundwater’s flow. Sampling
along linear transects helps establish the plume’s limits.
0 2 4 6 8 10
-1
.0
-0
.5
0.
0
0.
5
1.
0_{(a)}
(b)
Landfill
Direction of
groundwater flow

**StRatified Sampling**

Another combination of the three primary approaches to sampling is judg-mental–random, or stratified sampling. Many target populations consist of distinct units, or strata. For example, suppose we are studying particulate Pb in urban air. Because particulates come in a range of sizes—some visible and some microscopic—and from many sources—road dust, diesel soot, and fly ash to name a few—we can subdivide the target population by size or source. If we choose a random sampling plan, then we collect samples without considering the different strata. For a stratified sampling, we divide the target population into strata and collect random samples from within each stratum. After analyzing the samples from each stratum, we pool their respective means to give an overall mean for the target population. The advantage of stratified sampling is that individual strata usually are more homogeneous than the target population. The overall sampling variance for stratified sampling is always at least as good, and often better than that obtained by simple random sampling.

**convenience Sampling**

One additional method of sampling deserves brief mention. In conve-nience sampling we select sample sites using criteria other than minimiz-ing samplminimiz-ing error and samplminimiz-ing variance. In a survey of rural groundwater quality, for example, we can choose to drill wells at randomly selected sites or we can make use of existing wells, which is usually the preferred choice. In this case cost, expedience, and accessibility are more important than ensuring a random sample.

**7B.2 What Type of Sample to Collect**

After determining where to collect samples, the next step in designing a sampling plan is to decide what type of sample to collect. There are three common methods for obtaining samples: grab sampling, composite sam-pling, and in situ sampling.

The most common type of sample is a grab sample, in which we collect a portion of the target population at a specific time and/or location, provid-ing a “snapshot” of the target population. If our target population is homo-geneous, a series of random grab samples allows us to establish its properties. For a heterogeneous target population, systematic grab sampling allows us to characterize how its properties change over time and/or space.

A composite sample is a set of grab samples that we combine into a single sample before analysis. Because information is lost when we com-bine individual samples, we normally analyze grab sample separately. In some situations, however, there are advantages to working with a composite sample.

One situation where composite sampling is appropriate is when our interest is in the target population’s average composition over time or space.

For example, wastewater treatment plants must monitor and report the average daily composition of the treated water they release to the environ-ment. The analyst can collect and analyze individual grab samples using a systematic sampling plan, reporting the average result, or she can combine the grab samples into a single composite sample. Analyzing a single com-posite sample instead of many individual grab samples, saves time and money.

Composite sampling is also useful when a single sample can not sup-ply sufficient material for the analysis. For example, analytical methods for determining PCB’s in fish often require as much as 50 g of tissue, an amount that may be difficult to obtain from a single fish. By combining and homogenizing tissue samples from several fish, it is easy to obtain the necessary 50-g sample.

A significant disadvantage of grab samples and composite samples is that we cannot use them to continuously monitor a time-dependent change in the target population. In situ sampling, in which we insert an analytical sensor into the target population, allows us to continuously monitor the target population without removing individual grab samples. For example, we can monitor the pH of a solution moving through an industrial produc-tion line by immersing a pH electrode in the soluproduc-tion’s flow.

### Example 7.4

A study of the possible relationship between traffic density and the
con-centrations of lead, cadmium, and zinc in roadside soils, made use of the
following sampling plan.7_{ Samples of surface soil (0–10 cm) were collected }
at perpendicular distances of 1, 5, 10, 20, and 30 m from the roadway. At
each distance, 10 samples were taken from different locations and mixed
to form a single sample. What type of sampling plan is this? Explain why
this is an appropriate sampling plan.

**S**

**S**

**olution**This is an example of a systematic–judgemental sampling plan using com-posite samples. These are good choices given the goals of the study. Auto-mobile emissions release particulates containing elevated concentrations of lead, cadmium, and zinc—this study was conducted in Uganda where leaded gasoline is still in use—which settle out on the surrounding road-side soils as “dry rain.” Sampling in areas near roadways, and sampling at fixed distances from the roadway provides sufficient data for the study, while limiting the total number of samples. Combining samples from the same distance into a single, composite sample has the advantage of de-creasing sampling uncertainty. Because variations in metal concentrations perpendicular to the roadway is not of interest, the composite samples do not result in a loss of information.

**7B.3 How Much Sample to Collect**

To minimize sampling errors, samples must be of an appropriate size. If a sample is too small, its composition may differ substantially from that of the target population, introducing a sampling error. Samples that are too large, however, require more time and money to collect and analyze, with-out providing a significant improvement in the sampling error.

Let’s assume that our target population is a homogeneous mixture of
two types of particles. Particles of type A contain a fixed concentration of
analyte, and particles without analyte are of type B. Samples from this target
population follow a binomial distribution. If we collect a sample containing
*n particles, the expected number of particles containing analyte, n*A, is

*n*_{A} =*np*

where p is the probability of selecting a particle of type A. The standard deviation for sampling is

*s*_{samp}= *np*(1−*p*) 7.3

To calculate the relative standard deviation for sampling, *s*samprel , we divide

equation 7.3 by n_{A}, obtaining

*s* *np* *p*

*np*
samp

rel _{=} (1− )

Solving for n allows us to calculate the number of particles providing the desired relative sampling variance.

*n* *p*

*p* *s*

=1− × 1 _{2}

( _{samp}rel ) 7.4

### Example 7.5

Suppose we are analyzing a soil where the particles containing analyte represent only 1 × 10–7% of the population. How many particles must we collect to give a percent relative standard deviation for sampling of 1%?

**S**

**S**

**olution**Since the particles of interest account for 1 × 10–7% of all particles, the probability, p, of selecting one of these particles is only 1 × 10–9. Substitut-ing into equation 7.4 gives

*n*= − ×
× × = ×
−
−
1 1 10
1 10
1
0 01 1 10
9
9 2
13
( )
( . )

To obtain a relative standard deviation for sampling of 1%, we need to collect 1 × 1013 particles.

For a review of the binomial distribution, see Chapter 4.

A sample containing 1013 particles may be fairly large. Suppose this is equivalent to a mass of 80 g. Working with a sample this large clearly is not practical. Does this mean we must work with a smaller sample and accept a larger relative standard deviation for sampling? Fortunately the answer is no. An important feature of equation 7.4 is that the relative standard deviation for sampling is a function of the number of particles, not the combined mass of the particles. If we crush and grind the particles to make them smaller, then a sample containing 1013 particles will have a smaller mass. If we assume that a particle is spherical, then its mass is proportional to the cube of its radius.

mass∝* _{r}*3

Decreasing a particle’s radius by a factor of 2, for example, decreases its mass by a factor of 23, or 8.

### Example 7.6

Assume that a sample of 1013 particles from Example 7.5 weighs 80 g and that the particles are spherical. By how much must we reduce a particle’s radius if we wish to work with 0.6-g samples?

**S**

**S**

**olution**To reduce the sample’s mass from 80 g to 0.6 g, we must change its mass by a factor of

80

0 6. =133×

To accomplish this we must decrease a particle’s radius by a factor of

*r*
*r*
3 _{133}
5 1
= ×
= . ×

Decreasing the radius by a factor of approximately 5 allows us to decrease the sample’s mass from 80 g to 0.6 g.

Treating a population as though it contains only two types of particles
is a useful exercise because it shows us that we can improve the relative
standard deviation for sampling by collecting more particles. Of course, a
real population contains more than two types of particles, with the analyte
present at several levels of concentration. Nevertheless, many well-mixed
populations, in which the population’s composition is homogeneous on the
scale at which we sample, approximate binomial sampling statistics. Under
these conditions the following relationship between the mass of a random
grab sample, m, and the percent relative standard deviation for sampling,
*R, is often valid*

*mR*2_{=}*K*

s 7.5

where K_{s} is a sampling constant equal to the mass of a sample producing a
percent relative standard deviation for sampling of ±1%.8

### Example 7.7

The following data were obtained in a preliminary determination of the amount of inorganic ash in a breakfast cereal.

Mass of Cereal (g) 0.9956 0.9981 1.0036 0.9994 1.0067

% w/w Ash 1.34 1.29 1.32 1.26 1.28

What is the value of K_{s} and what size samples are needed to give a percent
relative standard deviation for sampling of ±2.0%. Predict the percent
relative standard deviation and the absolute standard deviation if we
col-lect 5.00-g samples.

**S**

**S**

**olution**To determine the sampling constant, Ks, we need to know the average mass of the cereal samples and the relative standard deviation for the amount of ash. The average mass of the cereal samples is 1.0007 g. The average % w/w ash and its absolute standard deviation are, respectively, 1.298% w/w and 0.03194% w/w. The percent relative standard deviation, R, therefore, is

*R* *s*
*X*
= samp× = w/w× =
w/w
100 0 03194
1 298 100 2 4
. %
. % . 66%

Solving for K_{s} gives its value as

*K*_{s} =*mR*2 =_{( .}_{1 0007}g_{)( . )}_{2 46} 2=_{6 06}_{.} g

To obtain a percent relative standard deviation of ±2%, samples need to have a mass of at least

*m* *K*
*R*
= s = =
2
g
(2.0) g
2
6 06. _{1 5}_{.}

If we use 5.00-g samples, then the expected percent relative standard
de-viation is
*R* *K*
*m*
= s = g =
5.00 g
6 06. _{1 10}_{. %}
and the expected absolute standard deviation is

*s*samp *RX*
w/w _{w}
= = =
100
1 10 1 298
100 0 0143
( . )( . % ) _{.} _{% //w}

8 Ingamells, C. O.; Switzer, P. *Talanta***1973**, *20*, 547–568.

Problem 8 in the end of chapter problems asks you to derive equation 7.5.

**7B.4 How Many Samples to Collect**

In the previous section we considered how much sample we need to mini-mize the standard deviation due to sampling. Another important consid-eration is the number of samples to collect. If our samples are normally distributed, then the confidence interval for the sampling error is

µ samp

samp
= ±*X* *ts*

*n* 7.6

where n_{samp} is the number of samples and s_{samp} is the standard deviation
for sampling. Rearranging equation 7.6 and substituting e for the quantity

*X* −µ, gives the number of samples as

*n* *t s*
*e*
samp
samp
=
2 2
2 7.7

Because the value of t depends on nsamp, the solution to equation 7.7 is found iteratively.

### Example 7.8

In Example 7.7 we determined that we need 1.5-g samples to establish an
*s*_{samp} of ±2.0% for the amount of inorganic ash in cereal. How many 1.5-g
samples do we need to obtain a percent relative sampling error of ±0.80%
at the 95% confidence level?

**S**

**S**

**olution**Because the value of t depends on the number of samples—a result we have yet to calculate—we begin by letting nsamp= ∞ and using t(0.05, ∞) for t. From Appendix 4, the value for t(0.05, ∞) is 1.960. Substituting known

values into equation 7.7 gives the number of samples as When we use equation 7.7, the standard

deviation for sampling, *s*_{samp}, and the
er-ror, *e*, must be expressed in the same way.
Because *s*_{samp} is given as a percent relative
standard deviation, the error, *e*, is given
as a percent relative error. When you use
equation 7.7, be sure to check that you are
expressing *s*_{samp} and *e* in the same way.

### Practice Exercise 7.3

Olaquindox is a synthetic growth promoter in medicated feeds for pigs. In an analysis of a production lot of feed, five samples with nominal masses of 0.95 g were collected and analyzed, with the results shown in the following table.

mass (g) 0.9530 0.9728 0.9660 0.9402 0.9576

mg olaquindox/kg feed 23.0 23.8 21.0 26.5 21.4
What is the value of *K*_{s} and what size samples are needed to obtain a
percent relative deviation for sampling of 5.0%? By how much do you
need to reduce the average particle size if samples must weigh no more
than 1 g?

*n*_{samp}=( . ) ( . ) = ≈
( . ) .
1 960 2 0
0 80 24 0 24
2 2
2

Letting *n*_{samp}= 24, the value of *t(0.05, 23) from *Appendix 4 is 2.073.
Recalculating nsamp gives

*n*_{samp} =( . ) ( . ) = ≈
( . ) .
2 073 2 0
0 80 26 9 27
2 2
2

When n_{samp}= 27, the value of t(0.05, 26) from Appendix 4 is 2.060.
Re-calculating n_{samp} gives

*n*_{samp}=( . ) ( . ) = ≈
( . ) .
2 060 2 0
0 80 26 52 27
2 2
2

Because two successive calculations give the same value for nsamp, we have an iterative solution to the problem. We need 27 samples to achieve a per-cent relative sampling error of ±0.80% at the 95% confidence level.

With 24 samples, the degrees of freedom
for *t* is 23.

### Practice Exercise 7.4

Assuming that the percent relative standard deviation for sampling in the determination of olaquindox in medicated feed is 5.0% (see Practice Ex-ercise 7.3), how many samples do we need to analyze to obtain a percent relative sampling error of ±2.5% at a = 0.05?

Click here to review your answer to this exercise.

Equation 7.7 provides an estimate for the smallest number of samples
that will produce the desired sampling error. The actual sampling error
may be substantially larger if *s*_{samp} for the samples we collect during the
subsequent analysis is greater than ssamp used to calculate nsamp. This is not
an uncommon problem. For a target population with a relative sampling
variance of 50 and a desired relative sampling error of ±5%, equation 7.7

predicts that 10 samples are sufficient. In a simulation using 1000 samples
of size 10, however, only 57% of the trials resulted in a sampling error of
less than ±5%.9_{ Increasing the number of samples to 17 was sufficient to }
ensure that the desired sampling error was achieved 95% of the time.

**7B.5 Minimizing the Overall Variance**

A final consideration when developing a sampling plan is to minimize the
overall variance for the analysis. Equation 7.2 shows that the overall
vari-ance is a function of the varivari-ance due to the method, *s*meth

2 _{, and the variance }

due to sampling, *s*_{samp}2 _{. As we have seen, we can improve the sampling }

vari-ance by collecting more samples of the proper size. Increasing the number

9 Blackwood, L. G. *Environ. Sci. Technol.***1991**, *25*, 1366–1367.

For an interesting discussion of why the number of samples is important, see Ka-plan, D.; Lacetera, N.; KaKa-plan, C. “Sample Size and Precision in NIH Peer Review,” Plos One, 2008, 3(7), 1–3. When review-ing grants, individual reviewers report a score between 1.0 and 5.0 (two significant figure). NIH reports the average score to three significant figures, implying that dif-ferences of 0.01 are significant. If the indi-vidual scores have a standard deviation of 0.1, then a difference of 0.01 is significant at a = 0.05 only if there are 384 reviews. The authors conclude that NIH review panels are too small to provide a statisti-cally meaningful separation between pro-posals receiving similar scores.

of times we analyze each sample improves the method’s variance. If *s*_{samp}2 _{ is }

significantly greater than *s*_{meth}2 _{, we can ignore the method’s contribution to }

the overall variance and use equation 7.7 to estimate the number of samples to analyze. Analyzing any sample more than once will not improve the overall variance, because the method’s variance is insignificant.

If *s*_{meth}2 _{ is significantly greater than }* _{s}*
samp

2 _{, then we need to collect and }

analyze only one sample. The number of replicate analyses, n_{rep}, needed to
minimize the error due to the method is given by an equation similar to

equation 7.7.
*n* *t s*
*e*
rep= meth
2 2
2

Unfortunately, the simple situations described above are often the ex-ception. For many analyses, both the sampling variance and the method variance are significant, and both multiple samples and replicate analyses of each sample are necessary. The overall error in this case is

*e t* *s*
*n*
*s*
*n n*
= samp +
samp
meth
samp rep
2 2
7.8
Equation 7.8 does not have a unique solution as different combinations of
*n*_{samp} and n_{rep} give the same overall error. How many samples we collect and
how many times we analyze each sample is determined by other concerns,
such as the cost of collecting and analyzing samples, and the amount of
available sample.

### Example 7.9

An analytical method has a percent relative sampling variance of 0.40% and a percent relative method variance of 0.070%. Evaluate the percent relative error (a = 0.05) if you collect 5 samples, analyzing each twice, and if you collect 2 samples, analyzing each 5 times.

**S**

**S**

**olution**Both sampling strategies require a total of 10 analyses. From Appendix 4

we find that the value of t(0.05, 9) is 2.262. Using equation 7.8, the rela-tive error for the first sampling strategy is

*e*= +
× =
2 262 0 40
5
0 070
5 2 0 67
. . . . %

and that for the second sampling strategy is

*e*= +
× =
2 262 0 40
2
0 070
2 5 1 0
. . . . %

Because the method variance is smaller than the sampling variance, we obtain a smaller relative error if we collect more samples, analyzing each fewer times.

### Practice Exercise 7.5

An analytical method has a percent relative sampling variance of 0.10% and a percent relative method variance of 0.20%. The cost of collecting a sample is $20 and the cost of analyzing a sample is $50. Propose a sam-pling strategy that provides a maximum percent relative error of ±0.50% (a = 0.05) and a maximum cost of $700.

Click here to review your answer to this exercise.

**7C Implementing the Sampling Plan**

Implementing a sampling plan normally involves three steps: physically removing the sample from its target population, preserving the sample, and preparing the sample for analysis. Except for in situ sampling, we analyze a sample after removing it from its target population. Because sampling exposes the target population to potential contamination, the sampling device must be inert and clean.

After removing a sample from its target population, there is a danger that it will undergo a chemical or physical change before we can complete its analysis. This is a serious problem because the sample’s properties no longer are representative of the target population. To prevent this problem, we often preserve samples before transporting them to the laboratory for analysis. Even when analyzing samples in the field, preservation may still be necessary.

The initial sample is called the primary or gross sample, and may be a single increment drawn from the target population, or a composite of sev-eral increments. In many cases we cannot analyze the gross sample without first reducing the sample’s particle size, converting the sample into a more readily analyzable form, or improving its homogeneity.

**7C.1 Solutions**

Typical examples of solution samples include those drawn from containers of commercial solvents; beverages, such as milk or fruit juice; natural waters, including lakes, streams, seawater and rain; bodily fluids, such as blood and urine; and, suspensions; such as those found in many oral medications. Let’s use the sampling of natural waters and wastewaters as a case study in how to sample solutions.

**Sample collection**

The chemical composition of a surface water—such as a stream, river, lake, estuary, or ocean—is influenced by flow rate and depth. Rapidly

flow-Although you may never work with the specific samples highlighted in this sec-tion, the case studies presented here may help you in envisioning potential prob-lems associated with your samples.

ing shallow streams and rivers, and shallow (<5 m) lakes are usually well mixed, and show little stratification with depth. To collect a grab sample we submerge a capped bottle below the surface, remove the cap and allow the bottle to fill completely, and replace the cap. Collecting a sample this way avoids the air–water interface, which may be enriched with heavy metals or contaminated with oil.10

Slowly moving streams and rivers, lakes deeper than five meters, estuar-ies, and oceans may show substantial stratification. Grab samples from near the surface are collected as described earlier, and samples at greater depths are collected using a sample bottle lowered to the desired depth (Figure 7.6).

Wells for sampling groundwater are purged before collecting samples because the chemical composition of water in the well-casing may be sig-nificantly different from that of the groundwater. These differences may result from contaminants introduced while drilling the well, or by a change in the groundwater’s redox potential following its exposure to atmospheric oxygen. In general, a well is purged by pumping out a volume of water equivalent to several well-casing volumes, or until the water’s temperature, pH, or specific conductance is constant. A municipal water supply, such as a residence or a business, is purged before sampling because the chemical composition of water standing in a pipe may differ significantly from the treated water supply. Samples are collected at faucets after flushing the pipes for 2-3 minutes.

Samples from municipal wastewater treatment plants and industrial discharges often are collected as a 24-hour composite. An automatic sam-pler periodically removes an individual grab sample, adding it to those col-lected previously. The volume of each sample and the frequency of sampling may be constant, or may vary in response to changes in flow rate.

Sample containers for collecting natural waters and wastewaters are made from glass or plastic. Kimax and Pyrex brand borosilicate glass have the advantage of being easy to sterilize, easy to clean, and inert to all solu-tions except those that are strongly alkaline. The disadvantages of glass containers are cost, weight, and the ease of breakage. Plastic containers are made from a variety of polymers, including polyethylene, polypropyl-ene, polycarbonate, polyvinyl chloride, and Teflon. Plastic containers are lightweight, durable, and, except for those manufactured from Teflon, in-expensive. In most cases glass or plastic bottles may be used interchange-ably, although polyethylene bottles are generally preferred because of their lower cost. Glass containers are always used when collecting samples for the analysis of pesticides, oil and grease, and organics because these species of-ten interact with plastic surfaces. Because glass surfaces easily adsorb metal ions, plastic bottles are preferred when collecting samples for the analysis of trace metals.

10 Duce, R. A.; Quinn, J. G. Olney, C. E.; Piotrowicz, S. R.; Ray, S. J.; Wade, T. L. *Science***1972**,

*176*, 161–163.

**Figure 7.6** A Niskin sampling
bottle for collecting water
sam-ples from lakes and oceans. After
lowering the bottle to the desired
depth, a weight is sent down the
winch line, tripping a spring that
closes the bottle. Source: NOAA

(photolib.noaa.gov).

### winch line

### spring

### cap

### cap

In most cases the sample bottle has a wide mouth, making it easy to fill and remove the sample. A narrow-mouth sample bottle is used if exposing the sample to the container’s cap or to the outside environment is a prob-lem. Unless exposure to plastic is a problem, caps for sample bottles are manufactured from polyethylene. When polyethylene must be avoided, the container cap includes an inert interior liner of neoprene or Teflon.

**Sample pReSeRvationand pRepaRation**

After removing a sample from its target population, its chemical composi-tion may change as a result of chemical, biological, or physical processes. To prevent a change in composition, samples are preserved by controlling the solution’s pH and temperature, by limiting its exposure to light or to the atmosphere, or by adding a chemical preservative. After preserving a sample, it may be safely stored for later analysis. The maximum holding time between preservation and analysis depends on the analyte’s stabil-ity and the effectiveness of sample preservation. Table 7.1 provides a list of representative methods for preserving samples and maximum holding times for several analytes of importance in the analysis of natural waters and wastewaters.

Most solution samples do not need additional preparation before analy-sis. This is the case for samples of natural waters and wastewaters. Solu-tion samples with particularly complex matricies—blood and milk are two examples—may need additional processing to separate the analytes from interferents, a topic covered later in this chapter.

**7C.2 Gases**

Typical examples of gaseous samples include automobile exhaust, emissions from industrial smokestacks, atmospheric gases, and compressed gases. Also

### Table 7.1 Preservation Methods and Maximum Holding Times for Selected

### Analytes in Natural Waters and Wastewaters

Analyte Preservation Method Maximum Holding Time

ammonia _{cool to 4 }o_{C; add H}

2SO4 to pH < 2 28 days

chloride none required 28 days

metals—Cr(VI) cool to 4 oC 24 hours

metals—Hg _{HNO}_{3}_{ to pH }_{< 2} 28 days

metals—all others _{HNO}_{3}_{ to pH }_{< 2} 6 months

nitrate none required 48 hours

organochlorine pesticides 1 mL of 10 mg/mL HgCl_{2} or
immediate extraction with a
suitable non-aqueous solvent

7 days without extraction 40 days with extraction

pH none required analyze immediately

Here our concern is only with the need to prepare the gross sample by converting it into a form suitable for analysis. Analytical methods may include additional sample preparation steps, such as concentrating or diluting the analyte, or adjusting the analyte’s chemical form. We will consider these forms of sample preparation in later chapters focusing on specific analytical methods.

included in this category are aerosol particulates—the fine solid particles and liquid droplets that form smoke and smog. Let’s use the sampling of urban air as a case study in how to sample gases.

**Sample collection**

One approach for collecting a sample of urban air is to fill a stainless steel canister or a Tedlar/Teflon bag. A pump pulls the air into the container and, after purging, the container is sealed. This method has the advantage of simplicity and of collecting a representative sample. Disadvantages in-clude the tendency for some analytes to adsorb to the container’s walls, the presence of analytes at concentrations too low to detect with accuracy and precision, and the presence of reactive analytes, such as ozone and nitrogen oxides, that may react with the container or that may otherwise alter the sample’s chemical composition during storage. When using a stainless steel canister, cryogenic cooling, which changes the sample from a gaseous state to a liquid state, may limit some of these disadvantages.

Most urban air samples are collected using a trap containing a solid sorbent or by filtering. Solid sorbents are used for volatile gases (vapor pres-sures more than 10–6 atm) and semi-volatile gases (vapor pressures between 10–6 atm and 10–12 atm). Filtration is used to collect aerosol particulates. Trapping and filtering allows for sampling larger volumes of gas—an im-portant concern for an analyte with a small concentration—and stabilizes the sample between its collection and its analysis.

In solid sorbent sampling, a pump pulls the urban air through a can-ister packed with sorbent particles. Typically 2–100 L of air are sampled when collecting volatile compounds, and 2–500 m3 when collecting semi-volatile gases. A variety of inorganic, organic polymer, and carbon sorbents have been used. Inorganic sorbents, such as silica gel, alumina, magnesium aluminum silicate, and molecular sieves, are efficient collectors for polar compounds. Their efficiency for collecting water, however, limits their sorp-tion capacity for many organic compounds.

Organic polymeric sorbents include polymeric resins of
*2,4-diphenyl-p-phenylene oxide or styrene-divinylbenzene for volatile compounds, and *
polyurethane foam for semi-volatile compounds. These materials have a
low affinity for water, and are efficient collectors for all but the most highly
volatile organic compounds, and some lower molecular weight alcohols
and ketones. The adsorbing ability of carbon sorbents is superior to that of
organic polymer resins, which makes them useful for highly volatile organic
compounds that can not be collected by polymeric resins. The adsorbing
ability of carbon sorbents may be a disadvantage, however, since the sorbed
compounds may be difficult to desorb.

Non-volatile compounds are normally present either as solid particu-lates, or are bound to solid particulates. Samples are collected by pulling large volumes of urban air through a filtering unit, collecting the particu-lates on glass fiber filters.

The short term exposure of humans, animals, and plants to atmospheric
pollutants is more severe than that for pollutants in other matrices. Because
the composition of atmospheric gases can vary significantly over a time, the
continuous monitoring of atmospheric gases such as O_{3}, CO, SO_{2}, NH_{3},
H_{2}O_{2}, and NO_{2} by in situ sampling is important.11

**Sample pReSeRvationand pRepaRation**

After collecting a gross sample of urban air, there is generally little need for sample preservation or preparation. The chemical composition of a gas sample is usually stable when it is collected using a solid sorbent, a filter, or by cryogenic cooling. When using a solid sorbent, gaseous compounds are released for analysis by thermal desorption or by extracting with a suitable solvent. If the sorbent is selective for a single analyte, the increase in the sor-bent’s mass can be used to determine the amount of analyte in the sample.

**7C.3 Solids**

Typical examples of solid samples include large particulates, such as those found in ores; smaller particulates, such as soils and sediments; tablets, pel-lets, and capsules used for dispensing pharmaceutical products and ani-mal feeds; sheet materials, such as polymers and rolled metals; and tissue samples from biological specimens. Solids are usually heterogeneous and samples must be collected carefully if they are to be representative of the target population. Let’s use the sampling of sediments, soils, and ores as a case study in how to sample solids.

**Sample collection**

Sediments from the bottom of streams, rivers, lakes, estuaries, and oceans are collected with a bottom grab sampler, or with a corer. A bottom grab sampler (Figure 7.7) is equipped with a pair of jaws that close when they contact the sediment, scooping up sediment in the process. Its principal advantages are ease of use and the ability to collect a large sample. Disad-vantages include the tendency to lose finer grain sediment particles as water flows out of the sampler, and the loss of spatial information—both laterally and with depth—due to mixing of the sample.

An alternative method for collecting sediments is a cylindrical coring device (Figure 7.8). The corer is dropped into the sediment, collecting a column of sediment and the water in contact with the sediment. With the possible exception of sediment at the surface, which may experience mix-ing, samples collected with a corer maintain their vertical profile, preserving information about how the sediment’s composition changes with depth.

Collecting soil samples at depths of up to 30 cm is easily accomplished with a scoop or shovel, although the sampling variance is generally high. A

11 Tanner, R. L. in Keith, L. H., ed. *Principles of Environmental Sampling*, American Chemical
Society: Washington, D. C., 1988, 275–286.

**Figure 7.7** Bottom grab sampler
being prepared for deployment.
Source: NOAA (photolib.noaa.
gov).

better tool for collecting soil samples near the surface is a soil punch, which is a thin-walled steel tube that retains a core sample after it is pushed into the soil and removed. Soil samples from depths greater than 30 cm are col-lected by digging a trench and collecting lateral samples with a soil punch. Alternatively, an auger may be used to drill a hole to the desired depth and the sample collected with a soil punch.

For particulate materials, particle size often determines the sampling
method. Larger particulate solids, such as ores, are sampled using a riffle
(Figure 7.9), which is a trough containing an even number of
compart-ments. Because adjoining compartments empty onto opposite sides of the
**Figure 7.8** Schematic diagram of a gravity corer in operation. The corer’s

weight is sufficient to penetrate into the sediment to a depth of approxi-mately 2 m. Flexible metal leaves on the bottom of the corer are pushed aside by the sediment, allowing it to enter the corer. The leaves bend back and hold the core sample in place as it is hauled back to the surface.

**Figure 7.9** Example of a four-unit riffle. Passing the
gross sample, shown within the circle, through the riffle
divides it into four piles, two on each side.
Combin-ing the piles from one side of the riffle provides a new
sample, which may be passed through the riffle again or
kept as the final sample. The piles from the other side of
the riffle are discarded.

gross sample

separated portions of sample

weight

core liner cable to ship

riffle, dumping a gross sample into the riffle divides it in half. By repeatedly passing half of the separated material back through the riffle, a sample of any desired size may be collected.

A sample thief (Figure 7.10) is used for sampling smaller particulate materials, such as powders. A typical sample thief consists of two tubes that are nestled together. Each tube has one or more slots aligned down the sample thief’s length. Before inserting the sample thief into the material being sampled, the slots are closed by rotating the inner tube. When the sample thief is in place, rotating the inner tube opens the slots, which fill with individual samples. The inner tube is then rotated to the closed posi-tion and the sample thief withdrawn.

**Sample pReSeRvation**

Without preservation, a solid sample may undergo a change in composition
due to the loss of volatile material, biodegradation, and chemical reactivity
(particularly redox reactions). Storing samples at lower temperatures makes
them less prone to biodegradation and to the loss of volatile material, but
fracturing of solids and phase separations may present problems. To
mini-mize the loss of volatiles, the sample container is filled completely,
eliminat-ing a headspace where gases collect. Samples that have not been exposed
to O_{2} are particularly susceptible to oxidation reactions. For example, the
contact of air with anaerobic sediments must be prevented.

**Sample pRepaRation**

Unlike gases and liquids, which generally require little sample preparation,
a solid sample usually needs some processing before analysis. There are two
reasons for this. First, as discussed in section 7B.3, the standard deviation
for sampling, ssamp, is a function of the number of particles in the sample,
not the combined mass of the particles. For a heterogeneous material
con-sisting of large particulates, the gross sample may be too large to analyze.
For example, a Ni-bearing ore with an average particle size of 5 mm may
require a sample weighing one ton to obtain a reasonable s_{samp}. Reducing
the sample’s average particle size allows us to collect the same number of
particles with a smaller, more manageable mass. Second, many analytical
techniques require that the analyte be in solution.

**Reducing paRticle Size**

A reduction in particle size is accomplished by a combination of crushing and grinding the gross sample. The resulting particulates are then thor-oughly mixed and divided into subsamples of smaller mass. This process seldom occurs in a single step. Instead, subsamples are cycled through the process several times until a final laboratory sample is obtained.

Crushing and grinding uses mechanical force to break larger particles into smaller particles. A variety of tools are used depending on the particle’s

**Figure 7.10** Sample thief showing
its open and closed positions.

open closed

size and hardness. Large particles are crushed using jaw crushers capable of reducing particles to diameters of a few millimeters. Ball mills, disk mills, and mortars and pestles are used to further reduce particle size.

A significant change in the gross sample’s composition may occur dur-ing crushdur-ing and grinddur-ing. Decreasdur-ing particle size increases the available surface area, which increases the risk of losing volatile components. This problem is made worse by the frictional heat that accompanies crushing and grinding. Increasing the surface area also exposes interior portions of the sample to the atmosphere where oxidation may alter the gross sample’s composition. Other problems include contamination from the materials used to crush and grind the sample, and differences in the ease with which particles are reduced in size. For example, softer particles are easier to reduce in size and may be lost as dust before the remaining sample is processed. This is a particular problem if the analyte’s distribution between different

types of particles is not uniform.

The gross sample is reduced to a uniform particle size by intermittently passing it through a sieve. Those particles not passing through the sieve receive additional processing until the entire sample is of uniform size. The resulting material is mixed thoroughly to ensure homogeneity and a sub-sample obtained with a riffle, or by coning and quartering. As shown in Figure 7.11, the gross sample is piled into a cone, flattened, and divided into four quarters. After discarding two diagonally opposed quarters, the remaining material is cycled through the process of coning and quartering until a suitable laboratory sample remains.

**Figure 7.11** Illustration showing the method of coning and quartering for reducing sample size. After gathering
the gross sample into a cone, the cone is flattened, divided in half, and then divided into quarters. Two opposing
quarters are combined to form the laboratory sample, or the subsample is sent back through another cycle. The
two remaining quarters are discarded.

gather material into a cone flatten the cone

divide in half divide into quarters

discard

laboratory sample _{new}
cycle

**BRinging Solid SampleS into Solution**

If you are fortunate, your sample will easily dissolve in a suitable solvent, requiring no more effort than gently swirling and heating. Distilled water is usually the solvent of choice for inorganic salts, but organic solvents, such as methanol, chloroform, and toluene are useful for organic materials.

When a sample is difficult to dissolve, the next step is to try digesting it with an acid or a base. Table 7.2 lists several common acids and bases, and summarizes their use. Digestions are carried out in an open container, usually a beaker, using a hot-plate as a source of heat. The main advantage of an open-vessel digestion is cost because it requires no special equipment. Volatile reaction products, however, are lost, resulting in a determinate error

if they include the analyte.

Many digestions are now carried out in a closed container using mi-crowave radiation as the source of energy. Vessels for mimi-crowave digestion are manufactured using Teflon (or some other fluoropolymer) or fused silica. Both materials are thermally stable, chemically resistant, transparent to microwave radiation, and capable of withstanding elevated pressures. A typical microwave digestion vessel, as shown in Figure 7.12, consists of an insulated vessel body and a cap with a pressure relief valve. The vessels are placed in a microwave oven (typically 6–14 vessels can be accommodated) and microwave energy is controlled by monitoring the temperature or pres-sure within one of the vessels.

### Table 7.2 Acids and Bases Used for Digesting Samples

Solution Uses and Properties

HCl (37% w/w) _{• } _{dissolves metals more easily reduced than H}_{2}_{ (E }o_{<}_{ 0)}
dissolves insoluble carbonate, sulfides, phosphates, fluorides,
•

sulfates, and many oxides
HNO_{3} (70% w/w) • strong oxidizing agent

dissolves most common metals except Al, Au, Pt, and Cr •

decomposes organics and biological samples (wet ashing) •

H2SO4 (98% w/w) • dissolves many metals and alloys

decomposes organics by oxidation and dehydration •

HF (50% w/w) • dissolves silicates by forming volatile SiF_{4}

HClO_{4} (70% w/w) • hot, concentrated solutions are strong oxidizing agents
dissolves many metals and alloys

•

decomposes organics (

• *Caution: reactions with organics often are *
*explosive; use only in a specially equipped hood with a blast shield *
*and after prior decomposition with HNO3*)

HCl:HNO3 (3:1 v/v) • also known as aqua regia dissolves Au and Pt •

A microwave digestion has several important advantages over an open-vessel digestion, including higher temperatures (200–300 oC) and pres-sures (40–100 bar). As a result, digestions requiring several hours in an open-vessel may need no more than 30 minutes when using a microwave digestion. In addition, a closed container prevents the loss of volatile gases. Disadvantages include the inability to add reagents during the digestion, limitations on the sample’s size (typically < 1 g), and safety concerns due to the high pressures and corrosive reagents.

Inorganic samples that resist decomposition by digesting with acids or bases often can be brought into solution by fusing with a large excess of an alkali metal salt, called a flux. After mixing the sample and the flux in a crucible, they are heated to a molten state and allowed to cool slowly to room temperature. The melt usually dissolves readily in distilled water or dilute acid. Table 7.3 summarizes several common fluxes and their uses. Fusion works when other methods of decomposition do not because of the high temperature and the flux’s high concentration in the molten liquid.

### Table 7.3 Common Fluxes for Decomposing Inorganic Samples

Flux Temperature (Melting oC) Crucible Typical Samples

Na_{2}CO_{3} 851 Pt silicates, oxides, phosphates, sulfides

Li_{2}B_{4}O_{7} 930 Pt, graphite aluminosilicates, carbonates
LiBO_{2} 845 Pt, graphite aluminosilicates, carbonates

NaOH 318 Au, Ag silicates, silicon carbide

KOH 380 Au, Ag silicates, silicon carbide

Na_{2}O_{2} — Ni silicates, chromium steels, Pt alloys

K_{2}S_{2}O_{7} 300 Ni, porcelain oxides

B_{2}O_{3} 577 Pt silicates, oxides

digestion vessel pressure relief valve

### (b)

**Figure 7.12** Microwave digestion unit. (a) View of the unit’s interior showing the carousel holding the digestion
vessels. (b) Close-up of a Teflon digestion vessel, which is encased in a thermal sleeve. The pressure relief value,
which is part of the vessel’s cap, contains a membrane that ruptures if the internal pressure becomes too high.

Disadvantages include contamination from the flux and the crucible, and the loss of volatile materials.

Finally, we can decompose organic materials by dry ashing. In this
method the sample is placed in a suitable crucible and heated over a flame
or in a furnace. The carbon present in the sample oxidizes to CO_{2}, and
hydrogen, sulfur, and nitrogen leave as H2O, SO2, and N2. These gases can
be trapped and weighed to determine their concentration in the organic
material. Often the goal of dry ashing is to remove the organic material,
leaving behind an inorganic residue, or ash, that can be further analyzed.

**7D Separating the Analyte from Interferents**

When an analytical method is selective for the analyte, analyzing samples is a relatively simple task. For example, a quantitative analysis for glucose in honey is relatively easy to accomplish if the method is selective for glucose, even in the presence of other reducing sugars, such as fructose. Unfortu-nately, few analytical methods are selective toward a single species.

In the absence of interferents, the relationship between the sample’s
signal, S_{samp}, and the concentration of analyte, C_{A}, is

*S*samp=*k C*A A 7.9

where k_{A} is the analyte’s sensitivity. If an interferent, is present, then
equa-tion 7.9 becomes

*S*_{samp}=*k C*_{A A}+*kC*_{I I} _{7.10}

where k_{I} and C_{I} are, respectively, the interferent’s sensitivity and
concentra-tion. A method’s selectivity for the analyte is determined by the relative
difference in its sensitivity toward the analyte and the interferent. If *k*_{A}
is greater than kI, then the method is more selective for the analyte. The
method is more selective for the interferent if k_{I} is greater than k_{A}.

Even if a method is more selective for an interferent, we can use it to determine CA if the interferent’s contribution to Ssamp is insignificant. The

selectivity coefficient, *K*_{A,I}, which we introduced in Chapter 3,
pro-vides a way to characterize a method’s selectivity.

*K* *k*

*k*
A,I I

A

= _{7.11}

Solving equation 7.11 for kI, substituting into equation 7.10, and simplify-ing, gives

*S*_{samp}=*k C*_{A}( _{A}+*K*_{A,I}×*C*_{I}) _{7.12}
An interferent, therefore, does not pose a problem as long as the product of
its concentration and its selectivity coefficient is significantly smaller than
the analyte’s concentration.

In equation 7.9, and the equations that
follow, you can replace the analyte’s
con-centration, *C*_{A}, with the moles of analyte,

*n*_{A} when working with methods, such as
gravimetry, that respond to the absolute
amount of analyte in a sample. In this case
the interferent also is expressed in terms
of moles.

*K*_{A,I}× <<*C*_{I} *C*_{A}

If we cannot ignore an interferent’s contribution to the signal, then we must begin our analysis by separating the analyte and the interferent.

**7E General Theory of Separation Efficiency**

The goal of an analytical separation is to remove either the analyte or the interferent from the sample’s matrix. To achieve this separation there must be at least one significant difference between the analyte’s and the interfer-ent’s chemical or physical properties. A significant difference in proper-ties, however, may not be enough. A separation that completely removes the interferent may also remove a small amount of analyte. Altering the separation to minimize the analyte’s loss may prevent us from completely removing the interferent.

Two factors limit a separation’s efficiency—failing to recover all the analyte and failing to remove all the interferent. We define the analyte’s

recovery, R_{A}, as
*R* *C*
*C*
A A
A o
=
( ) 7.13

where *C*_{A} is the concentration of analyte remaining after the separation,
and (CA)o is the analyte’s initial concentration. A recovery of 1.00 means
that no analyte is lost during the separation. The interferent’s recovery, R_{I},
is defined in the same manner

*R* *C*
*C*
I I
I o
=
( ) 7.14

where C_{I} is the concentration of interferent remaining after the separation,
and (C_{I})_{o} is the interferent’s initial concentration. We define the extent of
the separation using a separation factor, S_{I,A}.12

*S* *R*

*R*
I,A I

A

= _{7.15}

In general, S_{I,A} should be approximately 10–7 for the quantitative analysis
of a trace analyte in the presence of a macro interferent, and 10–3 when the
analyte and interferent are present in approximately equal amounts.

### Example 7.10

An analytical method for determining Cu in industrial plating baths gives poor results in the presence of Zn. To evaluate a method for separating the analyte from the interferent, samples containing known

concentra-12 (a) Sandell, E. B. *Colorimetric Determination of Trace Metals*, Interscience Publishers: New York,
1950, pp. 19–20; (b) Sandell, E. B. * Anal. Chem. 1968*,

*40*, 834–835.

The meaning of trace and macro, as well as other terms for describing the concentra-tions of analytes and interferents, is pre-sented in Chapter 2.